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Thevenin

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qdn

New Member
I guess I should've worded it better. Does the dependent voltage source pumps its own current (not that of the 5A current source) into those 2 resistors? That is, if the dependent voltage source does pumps its own current into those 2 resistor, then the current pass through those resistors must be the sum of the current of both the 5A current source and the dependent voltage source which is not 5A.
 

The Electrician

Active Member
If we take the positive direction of current in the 4Ω and 6Ω resistors to be directed downward, then the sum of the currents in the two resistors due to the dependent source must be zero.

In other words, any current in the two resistors due to the dependent source must be equal in magnitude and opposite in direction.

The sum of the currents in the two resistors can't be anything but 5 amps. Only the 5 amp current source has one end connected to ground. The dependent source is floating, so it can't produce any net current with respect to ground.
 

hadiffina

New Member
Hi All,

Please help on the attached.

The question need to get the Thevenin voltage and resistance in between point AB.

Kindly advice,

Thanks & Best Regards,
Gilbert


1st step : find Rth

open 5A current source and introduce Vth between A and B, and
also Isc into the 2ohm resistor (which is series with Vth)

now there are 3 meshes, top mesh with I1, botom with I2 and on the right
bottom with I3. All currents are clockwise.
The equations are as below

Vx = -4I2 ; I3= -Isc

-2Vx + 2I1 -2I2 = 0
-2I1 + 12I2 - 6I3 = 0
0I1 - 6I2 + 8I3 = -Vth

solve these equations by matrix,
therefore I3 = -36Vth/216 = -Isc
( Vth/Isc ) = 216/36
Rth = 6ohm


2nd step : find Vth

placed back the 5A current source with terminal AB remains open
therefore Vth is equal voltage drop on the 6ohm resistor which denotes with V

This required supernode since there is an independent voltage source on
two essential nodes and the equation is as follow

-5 + (Vx/4) + [ (Vx-V) /2 ] + [ (V-Vx) /2 ] + V/6 = 0
therefore 3Vx + 2V = 60

-Vx - 2Vx + V = 0
Vx = (1/3)V

solving these equations will give V=20Volts
 
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